What is the probability of rolling the same number exactly three times

The probability of rolling {1, 1, 1, other, other} in this exact order is: \((\frac{1}{6})(\frac{1}{6})(\frac{1}{6})(\frac{5}{6})(\frac{5}{6}) =\frac{5^2}{6^5}\)
Now, the number of ways of rolling three 1s in 5 rolls is: \(C_5^3\) = 10
We must now account of all the numbers that can be the same (we could roll three 1s, three 2s, three 3s, three 4s, three 5s and three 6s). So we must multiply the number of possibilities by \(6\).

So then, we have: \((\frac{5^2}{6^5}) (10) (6) = \frac{125}{628}\)

D.

5 dices 6 side total outcomes= 6*6*6*6*6

favourable outcomes= (6c1*1*1*5c1*5c1)
but these places or same number and remaining two can be arranged in 5!/(3! 2!) ways

so total favourable outcomes = (6c1*1*1*5c1*5c1) * [5!/(3! 2!]

probability = favourable outcomes /total outcomes =125/648

Ans = D
_________________

We pick any number from 6.
The other two die need to show the same: (1/6 X 1/6)
Now we can pick the first number in 6 ways. Hence, ( 6 X 1/6 X 1/6)
Now the rest two die should not show the chosen number. ( 2 die X 5/6)
Altogether (6 X 1/6 X 1/6 X 2 X 5/6) = 5/18

First out of the six numbers(1,2,3,4,5,6) let us select one number that should be repeated on all the three dices
5C3=10
Now number of different ways to get three of the same number = 3!= 6
Now for the first dice to get the same number = 1/6 ways
For the second dice to get the same number we have = 1/6 ways
For the third dice to get the same number we have = 1/6 ways
The fourth dice can get any number except the which showed up on all the three dices = 5/6
The fourth dice can get any number except the which showed up on all the three dices = 5/6
So total prob = 10 x 6 x 1/6 x 1/6 x 1/6 x 5/6 x 5/6 = 125/648

Can you explain why you multiply by 5! / (3! * 2!)?

Also 5C3 that yields "10" but when I list out all the combinations, I get 15. Example below (with the number 1 being the chosen number)
11122 11123 11135
11133 11124 11136
11144 11125 11145
11155 11126 11146
11166 11134 11156

I only get 10 when you use the convention XXXNN where "X" is the number and "N" is any number not "X". I'm trying to find out where the disconnect is, especially since the 15 i listed above would be included in 6^5

Hey,
Could you clarify.....if (other,other)is of same kind or different kind?
The answer you have solved above will be true in case of {1,1,1,5,5} but not in case of{1,1,1,5,4}....
here,the number of ways of rolling three 1s in 5 rolls is more than 10....(you could check it manually)

Please clear my doubt...

Hi,

It really doesn't matter what the other two rolls are, as long as they're not 1. The first calculation I made above is for the very specific case where the first three rolls are 1 and the last two rolls are NOT 1. I calculated the probably of this case (that the first three rolls MUST BE 1 and the other two rolls can be any of 2, 3, 4, 5, or 6 in any combination) to be \(\frac{25}{6^5}\), which is just saying there are \(25\) combinations of {1,1,1,other,other} and \(6^5\) total possibilities (rolling 5 dice, each having 6 sides). Here is where I think you're getting turned around. The 25 combinations are given by:

{1,1,1,2,2} {1,1,1,2,3} {1,1,1,2,4} {1,1,1,2,5} {1,1,1,2,6}
{1,1,1,3,2} {1,1,1,3,3} {1,1,1,3,4} {1,1,1,3,5} {1,1,1,3,6}
{1,1,1,4,2} {1,1,1,4,3} {1,1,1,4,4} {1,1,1,4,5} {1,1,1,4,6}
{1,1,1,5,2} {1,1,1,5,3} {1,1,1,5,4} {1,1,1,5,5} {1,1,1,5,6}
{1,1,1,6,2} {1,1,1,6,3} {1,1,1,6,4} {1,1,1,6,5} {1,1,1,6,6}

But you have to remember that up to this point, we have considered only one triple-1 case: the case where the first three rolls are all 1. Here is where \(C_5^3\) comes into play. We would like to determine all the different combinations of having three 1s in five rolls, not just the first three rolls. Here, we have \(C_5^3 = 10\) combinations:

{1,1,1,-,-} {1,1,-,1,-} {1,1,-,-,1}
{1,-,1,1,-} {1,-,1,-,1} {1,-,-,1,1}
{-,1,1,1,-} {-,1,1,-,1} {-,1,-,1,1}
{-,-,1,1,1}

Finally, we have to account for the the other numbers that can be the same, not just 1. Hence, we multiply by 6.

I'm far from an expert but this process makes sense to me. Maybe someone else can put it more clearly.

VERITAS PREP OFFICIAL SOLUTION

Rolling any number on a dice three times in a row is equal to the number of throws 1/6^3*6 = 1/36, where 3 represents the number of throws and 6 is the number of different ways to get three of the same number (e.g. 1, 1, 1 2, 2, 2 3, 3, 3…).

Next we have to use the combinations formula to determine how many ways three out of five can be the same. 5!/(3!*2!)=10. We want to multiply this number to 1/36*10=10/36

We are not finished yet – there is one little twist to the problem. The question says “exactly three times.” Meaning we have to discount all the instances in which we roll the same number four times and those instance in which we roll the same number five times.

One way to do so is by multiplying by the probability that the fourth and fifth dice will NOT land on the same number as the three dice. Because there are five other possibilities on dice that will not compromise “exactly three numbers” we can multiply by 5/6*5/6=25/36

Finally this gives us 10/36*25/36= 125/648

Answer: D.
_________________

Total no. of possible cases = 6^5

Now to freeze three places where same digit will occur, no. of cases = 6C3
we can have any of the 6 digits on these 3 places, so no. of cases = 6C3 * 6
remaining 2 places can have any of the 5 digits (digit except that which is present on those 3 places) = 6C3 * 6 * (5)^2

probability = (6C3*6*(5)^2)/ (6^5) = 125/648

Answer D

I solved this using the formula

5C3* (1/6) ^3 * (5/6) ^2

Option D

For e.g. on a six-sided dice: P(getting 6) = \(\frac{1}{6}\) & P(not getting) = \(1 - \frac{1}{6} = \frac{5}{6}\)

P(getting three 6's on 5 dices) = \(5C3*(\frac{1}{6})^3*(\frac{5}{6})^2\) = \(\frac{250}{7776}\) = \(\frac{125}{3888}\)

Similarly, P(getting three 5's on 5 dices) = \(\frac{125}{3888}\)
P(getting three 4's on 5 dices) = \(\frac{125}{3888}\)
P(getting three 3's on 5 dices) = \(\frac{125}{3888}\)
P(getting three 2's on 5 dices) = \(\frac{125}{3888}\)
P(getting three 1's on 5 dices) = \(\frac{125}{3888}\)

P(getting same number exactly three times on 5 dices) = \(6*\frac{125}{3888}\) = \(\frac{125}{648}\)

Bunuel
You sure (D) is correct?
The question says that a number must occur exactly thrice.

The Original solution takes into account the favorable cases as

6C1(the number that will occur thrice) * 5C3(choosing the three positions for that number ) *5 * 5(the other two numbers)

But in this case, the other two numbers can be the same, and according to the question a number has to occur exactly thrice.

Shouldn't it be

6C1*5C3*5*4 ?

or maybe I'm missing something.

We have 5 throws of the dice.

We need EXACTLY 3 of the dice to show the same number and the remaining 2 dice to show a different number.

We can call the 3 numbers “same” = S
And
We can call the 2 numbers “different” = D

1st) Calculate one possibility

Since the 3 “same” dice can show up in any one of the 5 throws, we need to shuffle around (Arrange) the different ways we can have a successful outcome.

First, though, we can find a representative case.

S - S - S - D - D

The key is that it says Exactly 3 are the same —— so we could have 3 dice the same and the other 2 as a different “same” number, so long as they are not the same as the first 3 dice.

1st die thrown: can be any number: 1 or 2 or 3 or 4 or 5 or 6 —— Prob = 6/6

2nd die thrown: must be identical to the 1st die that appeared, so only 1 number showing will be considered a Favorable Outcome —- Prob = 1/6

3rd die thrown: same reasoning as the 2nd die thrown —— Prob = 1/6

4th die thrown: the 4th die thrown can be ANY NUMBER other than the one that appeared for the first 3 ——— gives us 5 favorable options out of 6 total —— Prob = 5/6

5th die thrown: same reasoning as the 4th die thrown. Prob = 5/6

So to get:

(Same) - (same) - (same) -(different) - (different)

In that exact order, the probability is:

(6/6) (1/6) (1/6) (5/6) (5/6) = (5)^2 / (6)^4

2nd) Binomial Probability Concept

We need to determine the different ways we can have 3 die be the “same” and 2 die be “different”

Essentially, we need to choose which 3 dice will be the “same” dice

5 c 3 =. 10 ways

Gives us:

(10) (5)^2
_________
(6)^4

Cancel the factor of 2 that appears in the NUM and DEN and we are left with:

(5) (5)^2
_________
(3) (6)^3


125
_______
(216) (3)

125/648

Answer *D*

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This troubled me for a bit as well.

Ultimately, a number must occur exactly thrice means we can have an outcome such as:

3-3-3-1-6

or

3-3-3-2-2

The other two can be the same as well.

If you read the official solution, the only cases that are eliminated are the ones in which the 4th dice and 5th dice land on the same number as the other 3 numbers that are exactly the same.

In other words, the cases in which 3 are the same and the other 2 are a DIFFERENT “same” are allowed.

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